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In a p-n junction, a potential barrier of 250 meV exists across the junction. A hole with a kinetic energy of 300 meV approaches the junction. Find the kinetic energy of the hole when it crosses the junction if the hole approached the junction (a) from the p-side and (b) from the n-side. |
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Answer» Potential barrier ‘d’ = 250 meV Initial KE of hole = 300 meV We know: KE of the hole decreases when the junction is forward biased and increases when reverse blased in the given ‘Pn’ diode. So, a) Final KE = (300 – 250) meV = 50 meV b) Initial KE = (300 + 250) meV = 550 meV |
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